It does not only look like it, it is the reason:
If ##\sigma## notes the cyclic shift by one digit (##\sigma(ABCDEF)=BCDEFA##) we get with ##x=ABCDEF##
$$
x \cdot 10^k \equiv \sigma^k(x) \operatorname{mod}7
$$
i.e. ##\sigma## acts like the multiplication by ##10## in ##\mathbb{Z}_7\,,## ##10:7=1.42857 \ldots \,.##

I tried the 1000! problem with logic, by counting the x10 and x100 numbers, then counting the x2 . x5 combinations that produce 1, 2, 3 and 4 zeroes... but I got the wrong total of zeroes compared to the real number from Wolfram-Alpha's ...

I tried the 1000! problem with logic, by counting the x10 and x100 numbers, then counting the x2 . x5 combinations that produce 1, 2, 3 and 4 zeroes... but I got the wrong total of zeroes compared to the real number from Wolfram-Alpha's ...

Because 2.5 = 10, so each group of 1.2.3.4.5.6.7.8.9 = 2.5.(1.3.4.6.7.8.9) = 10.(36,288)

So numbers that end in 1,3,4,6,7,8,9 don't add any zeroes to the result ( ex x3.x6 = (10x+3).(10x+6) = 100x^2 + 10.(3+6).x + 6.3... no zeroes in any x6.x3 for all x in {0..99} ).

As any number ending in 2 will be followed by a number ending in 5 in the sequence {1..1000}, then for x in 0..99 we have x2.x5 = 100x^2 + 10.(2+5)x + 10 = 100x^2+10(7x+1), therefore x2.x5 add zeroes. So we get 1x3 zeroes from 1000, 9x2=18 zeroes from 100, 200, .. , 900, 90x1 zeroes from 10, 20, ..., 90, 110, 120, ..., 190, 210, ... , 290, .. , 990, = 111 zeroes, plus the zeroes from the multiplications of x2.x5.

The zeroes for x2.x5 will be 1 for each x in 0..99... which is 100 numbers (ex 22.25 = 550); then 1 more for each (10.x^2 mod 10 + 7x+1) that is multiple of 10; then 1 more for each (x^2 mod 10 + 7x+1 ) that is multiple of 100. Inspection shows that only x=7, 17, 27, ..., 97 produces that expression as multiple of 10 (ex 372.375 = 139,500), and only 87 produces an expression that is multiple of 100 (872.875 = 736,000). So the x2.x5 terms produce 100 + 10 + 1 = 111 zeroes.

This calculation yields 222 zeroes, but Wolfram-Alpha shows 249 zeroes, so my result is wrong... I'm missing something and I have no clue what... :-(

Because 2.5 = 10, so each group of 1.2.3.4.5.6.7.8.9 = 2.5.(1.3.4.6.7.8.9) = 10.(36,288)

So numbers that end in 1,3,4,6,7,8,9 don't add any zeroes to the result ( ex x3.x6 = (10x+3).(10x+6) = 100x^2 + 10.(3+6).x + 6.3... no zeroes in any x6.x3 for all x in {0..99} ).

As any number ending in 2 will be followed by a number ending in 5 in the sequence {1..1000}, then for x in 0..99 we have x2.x5 = 100x^2 + 10.(2+5)x + 10 = 100x^2+10(7x+1), therefore x2.x5 add zeroes. So we get 1x3 zeroes from 1000, 9x2=18 zeroes from 100, 200, .. , 900, 90x1 zeroes from 10, 20, ..., 90, 110, 120, ..., 190, 210, ... , 290, .. , 990, = 111 zeroes, plus the zeroes from the multiplications of x2.x5.

The zeroes for x2.x5 will be 1 for each x in 0..99... which is 100 numbers (ex 22.25 = 550); then 1 more for each (10.x^2 mod 10 + 7x+1) that is multiple of 10; then 1 more for each (x^2 mod 10 + 7x+1 ) that is multiple of 100. Inspection shows that only x=7, 17, 27, ..., 97 produces that expression as multiple of 10 (ex 372.375 = 139,500), and only 87 produces an expression that is multiple of 100 (872.875 = 736,000). So the x2.x5 terms produce 100 + 10 + 1 = 111 zeroes.

This calculation yields 222 zeroes, but Wolfram-Alpha shows 249 zeroes, so my result is wrong... I'm missing something and I have no clue what... :-(